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I'm guessing that the answer to this question is well-known, but I'm struggling to find anything to help me.

Let $X,Y$ be compact manifolds of dimension $n,m$ respectively. Let $f:X \to Y$ be a smooth map. Then one can consider the graph $\Delta_f$ of $f$ as a cycle in $X \times Y$.

Firstly what is "known" about $\Delta_f$ considered as a homology class? (I appreciate that this is a little vague). There might need to be some extra conditions placed of $f$, as clearly if for example $f$ maps everything to a point then there is nothing to be said.

Secondly (and related to the first question), suppose that $n=m$, and $f$ is an immersion. Then the self intersection $\Delta_f^2$of $\Delta_f$ with itself is well-defined. Is there a simple expression for this in terms of the basic properties of $f$?

Thanks!

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If $n=m$ and $f$ is an immersion, then it is a covering map. For the 1-sheet covering the self-intersection number equals the Euler characteristic of $Y$, by Lefshetz theorem. In the general case, just multiply by the number of sheets. –  Sergei Ivanov Apr 29 '10 at 18:58

2 Answers 2

up vote 5 down vote accepted

A little late to the party, but here's what I think happens.

I will assume $X$ is closed and oriented, with fundamental class $[X]\in H_n(X)$. I will also assume we're using field coefficients (or $X$ has torsion-free homology) so there is an isomorphism $H_*(X)\otimes H_*(X)\cong H_*(X\times X)$ given by cross product.

Let $F\colon X\to X\times Y$ be the map $F(x)=(x,f(x))$. Then $\Delta_f=F_*[X]\in H_n(X\times Y)$. Note that there is a factorisation $F=(1\times f)\circ d$, where $d\colon X\to X\times X$ is the diagonal map.

Under our assumptions, a basis $\{b_i\}_{i\in I}$ for $H_*(X)$ has a Poincaré dual basis $\{b_i'\}$, and the diagonal class is given by $d_*[X]=\sum_{i\in I} b_i\times b_i'$ (see Chapter 11 of Milnor and Stasheff). Hence \[ \Delta_f = (1\times f)_*d_*[X] = \sum_{i\in I} b_i \times f_*(b_i'). \] So it's not enough to know just the homology class represented by $f$, you have to know the whole induced homology homomorphism.

With regards your second question, as noted in the comments, $f$ is a covering map. The map $F\colon X\to X\times Y$ is an embedding, whose normal bundle is isomorphic to the tangent bundle of $X$. So this self-intersection number $\Delta_f^2$ should just be the Euler characteristic of $X$.

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Well, $X$ was what I meant to write! This would agree with Sergei Ivanov's comment that $\Delta_f^2=n\chi(Y)=\chi(X)$ when $f$ has $n$ sheets. –  Mark Grant Dec 31 '10 at 20:24
    
Uh, yes. You are right. –  Sándor Kovács Dec 31 '10 at 23:24

We have a short exact sequence: $$ 0\to T_{\Delta_f} \to T_{X\times Y}|_{\Delta_f} \to N_{\Delta_f/Y} \to 0 $$ We also have that $T_{X\times Y}\simeq p_X^*T_X\oplus p_Y^*T_Y$ (where $p_X$ and $p_Y$ are the projections) and if $\Gamma_f: X\to X\times Y$ is the induced embedding, then $T_{\Delta_f}\simeq \Gamma_f^*T_X$.

This implies that then $$ {\rm deg}\ N_{\Delta_f/Y}={\rm deg}\ (p_Y^*T_Y)|_{\Delta_f} $$ and hence (in the case $m=n$) $$ \Delta_f^2={\rm deg} \left((p_Y^*T_Y)|_{\Delta_f}\right) = {\rm deg} \left(f^*T_Y\right) = {\rm deg} (f^*c_1(Y))= {\rm deg}\ f\cdot {\rm deg}\ c_1(Y). $$

By the way, I would not say that "if for example f maps everything to a point then there is nothing to be said". Granted, it is simple, but one can still say that, and also that $\Delta_f^2=0$.

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